Question: Solve the equation. $\dfrac32 + b = \dfrac74$ $b=$
Solution: Let's subtract to get $b$ by itself. $\begin{aligned} \dfrac32 + b &= \dfrac74 \\ \\ \dfrac32 +b {-\dfrac32}&= \dfrac74{-\dfrac32}~~~~~{\text{subtract }\dfrac32} \text{ from each side to get } b \text{ by itself }\\ \\ \cancel{ \dfrac32} +b {{-}\cancel{{\dfrac32}}}&= \dfrac74{-\dfrac32}\\ \\ b &= \dfrac74{-\dfrac32} \end{aligned}$ $\begin{aligned} \dfrac74-\dfrac32 &=\dfrac{7}{4}-\dfrac{3\times2}{2\times2}\\\\ &= \dfrac{7}{4}-\dfrac{6}{4} \\\\ &= \dfrac{1}{4} \end{aligned}$ The answer: $b={\dfrac14}$ Let's check to make sure. $\begin{aligned} \dfrac32 + b &= \dfrac74\\\\ \dfrac32 + {\dfrac14}&\stackrel{?}{=} \dfrac74 \\\\ \dfrac64 + {\dfrac14}&\stackrel{?}{=} \dfrac74 \\\\ \dfrac74 &= \dfrac74 ~~~~~~~~~~\text{Yes!} \end{aligned}$